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A sequence of numbers $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 3$ $a_{n+1} = 2a_n - c \quad (n > 1)$ where $c$ is a constant - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 2

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A-sequence-of-numbers-$a_1,-a_2,-a_3,-\ldots$-is-defined-by--$a_1-=-3$---$a_{n+1}-=-2a_n---c-\quad-(n->-1)$---where-$c$-is-a-constant-Edexcel-A-Level Maths Pure-Question 7-2012-Paper 2.png

A sequence of numbers $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 3$ $a_{n+1} = 2a_n - c \quad (n > 1)$ where $c$ is a constant. (a) Write down an expression,... show full transcript

Worked Solution & Example Answer:A sequence of numbers $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 3$ $a_{n+1} = 2a_n - c \quad (n > 1)$ where $c$ is a constant - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 2

Step 1

Write down an expression, in terms of $c$, for $a_2$

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Answer

To find a2a_2, we use the recurrence relation:

a2=2a1ca_2 = 2a_1 - c Substituting a1=3a_1 = 3 gives: a2=2(3)c=6c.a_2 = 2(3) - c = 6 - c.

Step 2

Show that $a_3 = 12 - 3c$

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Answer

Using the recurrence relation again to find a3a_3:

a3=2a2c.a_3 = 2a_2 - c. Substituting a2=6ca_2 = 6 - c:

a3=2(6c)c=122cc=123c.a_3 = 2(6 - c) - c = 12 - 2c - c = 12 - 3c.

Step 3

Given that $\sum_{i=1}^4 a_i \geq 23$, find the range of values of $c$.

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Answer

First, we compute a4a_4 using the recurrence relation:

a4=2a3c=2(123c)c=246cc=247c.a_4 = 2a_3 - c = 2(12 - 3c) - c = 24 - 6c - c = 24 - 7c. Then, we find the sum:

i=14ai=a1+a2+a3+a4=3+(6c)+(123c)+(247c)\sum_{i=1}^4 a_i = a_1 + a_2 + a_3 + a_4 = 3 + (6 - c) + (12 - 3c) + (24 - 7c)
=4511c.= 45 - 11c. Setting this greater than or equal to 23:

4511c2345 - 11c \geq 23
Solving for cc gives:

452311c2211cc2.45 - 23 \geq 11c \Rightarrow 22 \geq 11c \Rightarrow c \leq 2.
Thus, the range of values of cc that satisfies the condition is:

c2.c \leq 2.

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